Heights and Distances Practice Test 2

Question : 16 of 24

Marks: +1, -0

An observer measures angles of elevation of two towers of equal heights from a point between the towers. If the angles of elevation are 60° and 30° and distance of nearer tower is 100 m then the height of each tower and the distance between the towers, respectively are

$100/√3$ m and 300m
$100/√3$ m and 400m
$100 √3$ m and 300m
$100 √3$ m and 400m

Validate

Solution:

Let AB = CD = h m be the heights of the towers. E is a point such that DE = 100m; ∠CED = 60° and ∠AEB = 30°
Now, BE = x m (say)
From right triangle CDE,
h = 100 tan 60° =

100 √3

m
From right triangle ABE,
x = h cot 30° =

100√ 3×√ 3

= 300 m
Distance between the towers = DE + EB = 100 + 300 = 400 m
Height of the tower

= 100 √3

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